Tensor product: numerical example

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.4.

It’s useful to do a tensor calculation using actual numbers, just to see how they work. Suppose we have a vector with components

\displaystyle  A^{i}=\left[1,2,-1,0\right] \ \ \ \ \ (1)

where the units are metres, and a second vector given by

\displaystyle  B^{i}=\left[3,-1,0,-2\right] \ \ \ \ \ (2)

where the units are {\mbox{seconds}^{-1}}.

Since the units are different (even using relativistic units where length = time, since vector {B^{i}} has units of 1/time), we cannot add these vectors. However, we can multiply them to get a rank-2 tensor:

\displaystyle  M^{ij}=A^{i}B^{j}=\left[\begin{array}{cccc} 3 & -1 & 0 & -2\\ 6 & -2 & 0 & -4\\ -3 & 1 & 0 & 2\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (3)

The first row consists of {A^{0}} multiplied by each element of {B} in turn, and so on (the indices {i} and {j} each run for 0 to 3). The units of each element of {M^{ij}} are {\mbox{m}\cdot\mbox{s}^{-1}}.

The question in Moore asks if this product is commutative, which isn’t really the right thing to ask, since obviously {A^{i}B^{j}=B^{j}A^{i}}. However, the product is not symmetric, in the sense that {A^{i}B^{j}\ne A^{j}B^{i}}. That is, it matters which vector is used to determine the row index and which the column index.

The trace of {M} is given by

\displaystyle   M_{\;\; i}^{i} \displaystyle  = \displaystyle  g_{ij}M^{ij}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  g_{ij}A^{i}B^{j} \ \ \ \ \ (5)

This is just the scalar product {\mathbf{A}\cdot\mathbf{B}}. If we use the special relativity metric {\eta_{ij}}, then

\displaystyle  \eta_{ij}A^{i}B^{j}=-\left(1\times3\right)+\left(-1\times2\right)+\left(-1\times0\right)+\left(-2\times0\right)=-5 \ \ \ \ \ (6)

We could also have obtained this result from the matrix {M^{ij}} directly by calculating {\eta_{00}M^{00}+\eta_{11}M^{11}=-5} (the other two diagonal elements are zero).

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